# Simple poles and residues

1. Jan 8, 2009

### philip041

I have a function 1/(z^4 + 1)

I know the poles are pi/4, 3pi/4, 5pi/4, 7pi/4

It says they are simple poles, I thought I understood why, but now I am totally confused. How does my lecturer just know that they are simple?

A simple pole is a pole of order 1, but I thought this meant you look at the series and get the order from the highest indice of z on the denominator?

Help!

Cheers

2. Jan 8, 2009

### Count Iblis

The numerator has simple zeroes, so the function is pof the form:

1/[(z-a)(z-b)(z-c)(z-d)]

where a, b, c, and d are different. If you expand this around, say, z = a, you can proceed as follows. You can write the function as:

1/(z-a) * 1/[(z-b)(z-c)(z-d)]

The factor 1/[(z-b)(z-c)(z-d)] is not singular at z = a, so it has a regular Taylor expansion. So, the Laurent expansion of the function around z = a is given by the Taylor expansion of that factor times 1/(z-a).

3. Jan 8, 2009

### philip041

What is a simple zero?

4. Jan 8, 2009

### HallsofIvy

Staff Emeritus
Then you know wrong. The poles are $(\sqrt{2}/2)(1+ i)$, $-(\sqrt{2}/2)(1- i)$, $(-\sqrt{2}/2)(1+ i)$, and $(\sqrt{2}/2)(1- i)$, which can be written $e^{i\pi/4}$, $e^{3i\pi/4}$, $e^{5i\pi/4}$, and $e^{7i\pi/4}$.

$$\frac{1}{z^4+ 1}= \frac{1}{z- (\sqrt{2}/2)(1+ i)}\frac{1}{z+ (\sqrt{2}/2)(1-i)}\frac{1}{z+(\sqrt{2}/2)(1+ i)}\frac{1}{z-(\sqrt{2}/2)(1-i)}$$
Each of those numbers is a "simple pole" or "pole of order 1" because each has a power of 1 in the denominator.

5. Jan 8, 2009

### philip041

Cheers, this helps!